Math Olympiad! | Dare to solve this Puzzle! | AB=? |

[@jamestalbott4499]

Thank you! Enjoyed the problem solving using special triangles. Learning lots!

[@PreMath]

Great job! You are very welcome! Thanks ❤️

[@jimlocke9320]

At 8:40, we can apply the law of sines. Here, if the angles are designated by their vertices as A, B and C, and the sides opposite each angle by lower case a, b, and c, then a/sin(A) = b/sin(B) = c/sin(C). We use b/sin(B) = c/sin(C), for which we know b = 2, B = 45°, C = 105° and c = unknown AB. So, 2/sin(45°) = c/sin(105°). We derive sin(45°) from the special 45°-45°-90° triangle as 1/(√2). We note that sin(90° + Θ) = cos(Θ), so sin(105°) = cos(15°). We can derive cos(15°) from the known ratio of sides for a 15°-75°-90° triangle: 2 − √3 : 1 : √6 -√2, so cos(15°) = (2 − √3)/(√6 -√2), which simplifies to (√3 + 1)/(2√2). Alternatively, we can find the exact value of cos(15°) in a reference. So, 2/sin(45°) = c/sin(105°), 2/sin(45°) = c/cos(15°), 2/(1/√2) = c/((√3 + 1)/(2√2)), 2√2 = c/((√3 + 1)/(2√2)), and c = (2√2)((√3 + 1)/(2√2)) = √3 + 1. So, length AB = √3 + 1, as PreMath also found. While it was more complicated to divide ΔABC into 2 triangles, generate equations for each and solve, rather than apply the law of sines to ΔABC, PreMath's equations were much simpler to solve, which more than makes up for the extra complexity at the beginning!

[@PreMath]

Thanks ❤️

[@benben1049]

Drop an altitude from A to CD as AE. Set AE=a, because 45 right triangle, we have BE=a, Angel D is 30 degree, so DE=√3*a, BD=(√3-1)a, CE=[√3-2*(√3-1)] a, set up Pythagorean theorem on AEC to solve a, the a*√2 is x

[@PreMath]

Thanks ❤️

[@ybodoN]

Nice alternative giving AB = √(2 (2 + √3)) 😉

[@holyshit922]

It can be done with sine law and cosine law Let |CB| = |BD| = x |AB| = y Measures of angles in ABD Measure of angle A is 15 Measure of angle B is 135 Measure of angle D is 30 From sine law in ABD x/sin(15) = y/sin(30) x = y*sin(15)/sin(30) y = 2cos(15)*x From cosine law in ABC 4=4x^2cos^2(15)+x^2 - 2*x*2xcos(15)*sqrt(2)/2 4=x^2(4cos^2(15) + 1- 2sqrt(2)cos(15)) 4 = x^2 (2*(2cos^2(15) - 1)+3 - 2sqrt(2)*sqrt(2)/4(sqrt(3)+1)) 4 = x^2(2*cos(30)+3-sqrt(3)-1) 4 = x^2(2*sqrt(3)/2-sqrt(3)+2) 4 = 2x^2 x^2 - 2=0 x = sqrt(2) y = 2sqrt(2)*sqrt(2)/4*(sqrt(3)+1) y = sqrt(3)+1

[@LuisdeBritoCamacho]

As the triangle [ABC] is similar to triangle [ACD], although they are isomorphic triangles thy're not isometric ones. I can realize that angle CBA is equal to angle [CAD]. So angle [CAD] is equal to 45º - 15º = 30º and angle ACB = 105º. Using the Law of Sines we can state that: AB / sin(105º) = 2 / sin(45º) AB = 2*sin(105º) / sin(45º) AB = 1,932 / 0,707 ~ 2,73 li un. Answer: The segment AB is equal to approx. 2,73 linear units.

[@phungpham1725]

Let 's solve it inversely.😀 1/ Just build the special 30-90-60 triangle DEC first. We have: the triangle EBC is an equilateral one and the triangle EBD is an isosceles one of which each the 2 base angles = 30 degrees. 2/ Extend ED to the left a segment EA such that EA=EC----> the triangle AEC is a right isosceles so EA=EC=EB=BC----> the triangle AEB is an isosceles of which each of the base angle = 15 degrees (the exterior angle BDE=30 degrees). Now we have the required problem back😄 3/ Calculating: AE=CE=CB= sqrt2 ED= CE.sqrt3= sqrt2 .sqrt3----->AD= AE+ED= sqrt2 + sqrt2 .sqrt3= sqrt2(1+sqrt3) 4/ The angle CAB= 45-15= 30 degrees and the angle ABC= 60-15= 45 degrees which means that the 2 triangles ACB and DCA are similar: AB/AD=CB/CA---->AB/sqrt2(1+sqrt3) = sqrt2/2----> AB= 1+sqrt3 😄

[@yalchingedikgedik8007]

Thanks Sir Thanks PreMath That’s very nice ❤❤❤❤❤❤

[@PreMath]

Glad to hear that! You are very welcome! Thanks ❤️

[@marcgriselhubert3915]

As often, let's use an adapted orthonormal certer B (0;0) with A on the first axis A(-k; 0) with k>0. The equation of (BC) is y = -x as the angle ABC is 45° Let"s note C(-a;a) and its symetric D(a;-a). The angle BAD beeing 15°, the equation of (AD) is y = -tan(15°). (x +k) or y = (sqrt(3) -2). (a +k). As D is on AD) we have: -a = (sqrt(3) -2). (x+k) That gives a +k = a / (2 - sqrt(3)) = a. (2 +sqrt(3)) and k = a. (1 +sqrt(3)), So we have A(-a.(1 +sqrt(3); 0) and Vector AC(-a +a.(1 +sqrt(3)); a) or VectorAC (a.sqrt(3);a), giving that AC^2 =(a^2).3 + (a^2) = 4.(a^2) and that AC = 2.a As AC = 2 we can now say that a = 1 and that k = 1 + sqrt(3) = AB.

[@marcgriselhubert3915]

On line 3 please read: The angle BAD beeing 15°, the equation of (AD) is y = -tan(15°). (x + k) or y = (sqrt(3) -2). (x +k) As D is on (AD) we have: -a = (sqrt(3) -2). (a + k)

[@PreMath]

Thanks ❤️

[@MarieAnne.]

Here is a trig solution: Given: AC -= 2, BC = BD, ∠ABC = 45°, ∠BAD = 15° Therefore: ∠ABD = 135°, ∠ADB = 30° Let AB = x, BC = BD = y So we need to find x. Use Law of Sines in △ABD x/sin 30 = y/sin 15 x/(1/2) = y/[(√3−1)/(2√2)] y = (√3−1)x/√2 → y² = (4−2√3)x²/2 = (2−√3)x² Use Law of Cosines in △ABC cos 45° = (x² + y² − 2²) / (2xy) = √2/2 x² + y² − 4 = √2(xy) x² + (2−√3)x² − 4 = √2(x)(√3−1)x/√2 (3−√3)x² − 4 = (√3−1)x² (4−2√3)x² = 4 x² = 4/(4−2√3) = 4/(√3−1)² x = 2/(√3−1) = 2(√3+1)/((√3−1)(√3+1)) = 2(√3+1)/(3-1) *x = 1 + √3*

[@gelbkehlchen]

Solution: r = 1 = radius of the circle, a = side of the square. There is a right triangle with the hypotenuse r = 1 from the middle point of the left circle to the left edge point at the bottom from the sqare and the horizontal leg 1-a/2 and the vertical leg 1-a. Pythagoras: (1-a/2)²+(1-a)² = 1² ⟹ 1-a+a²/4+1-2a+a² = 1 |-1 ⟹ 5a²/4-3a+1 = 0 |*4/5 ⟹ a²-12/5*a+4/5 = 0 |p-q-formula ⟹ a1/2 = 6/5±√(36/25-4/5) = 6/5±√(36/25-20/25) = 6/5±√(16/25) = 6/5±4/5 ⟹ a1 = 6/5+4/5 = 10/5 = 2 [this square can’t be, it is to big] and a2 = 6/5-4/5 = 2/5 ⟹ solution.

[@PreMath]

Thanks ❤️

[@gelbkehlchen]

@@PreMath Excuse me. There is something wrong. This should be a solution to this task: kzbin.info/www/bejne/qJWUeIqGZ9KcqZI

[@mathbynisharsir5586]

Excellent explanation sir ❤❤❤❤❤

[@PreMath]

Thanks and welcome❤️

[@quigonkenny]

As ∠ABC = 45°, ∠DBA = 180-45 = 135°, as DC is a straight line, so the angles at B must add up to 180°. Since the internal angles of a triangle must add up to 180°, ∠ADB = 180-150 = 30°. Let E be the pont on AD where CE is perpendicular to AD. As ∠EDC = 30° and ∠CED = 90°, ∠DCE must equal 60°, and CE must equal DC/2. Therefore BC = CE, and ∆BCE is an equilateral triangle as (180-60)/2 = 60° and EB = BC. As ∠BED = 90-60 = 30°, and DB = BE, ∆DBE is an isosceles triangle and ∠DBE = 180-60 = 120°. Since ∠DBA = 135°, ∠EBA = 135-120 = 15°. As ∠BAE = ∠EBA = 15°, ∆AEB is isosceles and AE = EB. As AE = EC and ∠AEC is 90°, ∆AEC is also isosceles and ∠ECA = ∠CAE = (180-90)/2 = 45°. As ∠CAD = 45°, ∠DCA = 180-75 = 105°. By the Law of Sines, sin(a)/A = sin(b)/B where a and b are angles of a triangle and A and B are the lengths of their respective opposite sides. sin(45°)/2 = sin(105°)/AB AB(sin(45°)) = 2(sin(105°)) sin(x+y) = sin(x)cos(y) + cos(x)sin(y) sin(105) = sin(60)cos(45) + cos(60)sin(45) sin(105) = (√3/2)(√2/2) + (1/2)(√2/2) sin(105) = √6/4 + √2/4 = (√6+√2)/4 AB(√2/2) = 2(√6+√2)/4 AB = [(√6+√2)/2](2/√2) AB = (√6+√2)/√2 = √3 + 1 units

[@PreMath]

Thanks ❤️

[@prossvay8744]

Cos(105)=(√2)^2+2^2-AB^2/2(√2)(2) AB=2.73 units.❤❤❤ Thanks.sir.

[@PreMath]

Very good You are very welcome! Thanks ❤️

[@DB-lg5sq]

شكرا لكم على المجهودات يمكن استعمال AB=x BC=2xsin15 بتطبيق مبرهنة الكاشي على المثلث ABC x^2=4+2(3جذر) x=1+(3جذر)

[@PreMath]

Excellent! Thanks ❤️

[@ybodoN]

Alternative ending: If △CEA is 45° - 45° - 90° and △CED is 30° - 60° - 90° then CE = CB = BD = AE = √2 and ED = √2 · √3. Therefore AD = √2 (1 + √3). Then, by the AAA theorem, △ABC ~ △DAC and AB = 2√2 (1 + √3) / 2√2 which simplifies to (1 + √3) or approximately 2.73 units.

[@PreMath]

Thanks ❤️

[@surathpradhan1298]

Draw segment BE=BD on AD, and proof Angle C=90deg, i.e. AC=CB or AB=√8

[@PreMath]

Thanks ❤️

[@unknownidentity2846]

Let's do it: . .. ... .... ..... First of all we calculate some angles: ∠ABD = 180° − ∠ABC = 180° − 45° = 135° ∠ADB = 180° − ∠ABD − ∠BAD = 180° − 135° − 15° = 30° Let's call x the side length of BC and BD. Now we introduce point E on AD such that BE=x. Therefore BDE is an isosceles triangle and we can conclude: BD = BE ⇒ ∠BED = ∠BDE = ∠ADB = 30° ∠AEB = 180° − ∠BED = 180° − 30° = 150° ∠ABE = 180° − ∠AEB − ∠BAE = 180° − 150° − 15° = 15° So the triangle ABE is also an isosceles triangle: AE=BE=x. Now we can calculate according to the law of cosines: AB² = AE² + BE² − 2*AE*BE*cos(∠AEB) = x² + x² − 2*x²*cos(∠AEB) = 2*x²*[1 − cos(∠AEB)] = 2*x²*[1 − cos(150°)] = 2*x²*(1 + √3/2) = x²*(2 + √3) ⇒ AB = x*√(2 + √3) = x*(1 + √3)/√2 Now we apply the law of sines to the triangle ABC: ∠ACB = 180° − ∠ABC − ∠BAC = 180° − 45° − ∠BAC = 135° − ∠BAC AB/sin(∠ACB) = BC/sin(∠BAC) AB/BC = sin(∠ACB)/sin(∠BAC) [x*(1 + √3)/√2]/x = sin(135° − ∠BAC)/sin(∠BAC) (1 + √3)/√2 = [sin(135°)*cos(∠BAC) − cos(135°)*sin(∠BAC)]/sin(∠BAC) (1 + √3)/√2 = (1/√2)*[cos(∠BAC) + sin(∠BAC)]/sin(∠BAC) 1 + √3 = cos(∠BAC)/sin(∠BAC) + 1 √3 = cos(∠BAC)/sin(∠BAC) ⇒ ∠BAC = 30° Again we apply the law of sines to the triangle ABC to calculate AB: BC/sin(∠BAC) = AC/sin(∠ABC) x/sin(∠BAC) = AC/sin(∠ABC) ⇒ x = AC*sin(∠BAC)/sin(∠ABC) = 2*sin(30°)/sin(45°) = 2*(1/2)/(√2/2) = √2 ⇒ AB = x*(1 + √3)/√2 = √2*(1 + √3)/√2 = 1 + √3 Best regards from Germany

[@PreMath]

Great! Thanks ❤️

[@kinno1837]

Nice video!Btw do you have another method without using parallel lines to prove that angle sum of triangle is always equal to 180°?Because the setting of parallel lines is based on the sum of the interior angles of a triangle.Or is there any way to prove that alternating angles are equal in parallel lines?I look forward to your reply.

[@ybodoN]

You may want to read about the parallel postulate, also called Euclid's fifth postulate 😉

[@PreMath]

We may use the trigonometry! Thanks ❤️

[@prossvay8744]

Connect from B to E that angle BED=30° So BD=BE ; BDE is isoslated triangle and ABE=15°=EAB Triangle ABE is isosleted Hence: AE=BD=BE CBE=60° and BC=BE ត្រីកោណ CBE ជាត្រីកោណសម័ង្ស យើងបាន BC=BE=CE នាំអោយ មុំ BCE=CBE=BEC=60° មុំAEB=150° និង ភ្ជាប់់ C ទៅ E នាំអោយ AEC=150-60=90° និង AE=CE នាំអោយ ត្រីកោណ AEC ជាត្រីកោណកែងសមបាត. តាមទ្រឹស្តីបទពីតាគ័រ AC^2= CE^2+AE^2 4=2BC^2 BC=√2 មុំ CAE45° នាំអោយ មុំ BAC=30° គេបាន មុំ ACB=180°-30°-45°=105° តាមទ្រឹស្តីកូសុីនុស cos(105°)=(2^2+√2^2-AB^2)/(4√2)=2.73 . អរគុណ លោក គ្រូ.❤❤❤

[@PreMath]

Thanks ❤️

[@PlumbuM871]

Belissimo! 👍

[@PreMath]

Excellent! Thanks ❤️

[@StephenRayWesley]

(2)^2=4 3x(15°)=45°x 3x(15°)=45°x (90°+45°x+45°x)=180°x (180°x+2)=182°x (15°+15°)=30° (182°+30°)=√212°x √10^√20 2^√6 5^2 5^√4 √2^3^√2 √ 5√^2 √5^√2^2√ 1^3^√1√ 1^√1 √1^√1^2 32 (x+2x-3)

[@PreMath]

Thanks ❤️

[@giuseppemalaguti435]

Posto ACB=α..uso il teorema dei seni,con t=AB..2/sin45=t/sinα...a=CB...a/sin15=t/sin30..divido le due equazioni a=4(sin15/sin45)simα...l'altra equazione che lega a e α 2a/sin(30+α)=2/sin30…risolto risulta sinα=1/√(8-4√3)...α=75..infine t=2sinα/sin45=2√2sin75=2√(1+√3/2)=1+√3...boh,forse errori di distrazione

[@PreMath]

Thanks ❤️

[@yakupbuyankara5903]

AB=3^(1/2)+1

[@PreMath]

Thanks ❤️

[@wackojacko3962]

Yom! 🙂

[@PreMath]

Thanks ❤️

[@SimsimaFo]

CAN YOU DO MOROCCAN OLYMPIAD PLS

[@PreMath]

Will try! Thanks for asking❤️

[@Train_tv_22]

😂❤

[@PreMath]

Thanks ❤️

[@축복이-x6u]

asnwer=12cm

[@Train_tv_22]

Mujy math bilkul acha ni lgta tha